$h(t) = 5t^{2}+3t+1-5(f(t))$ $g(n) = -4n^{3}-6n^{2}+5(f(n))$ $f(t) = -4t^{2}-6t$ $ h(f(-1)) = {?} $
First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = -4(-1)^{2}+(-6)(-1)$ $f(-1) = 2$ Now we know that $f(-1) = 2$ . Let's solve for $h(f(-1))$ , which is $h(2)$ $h(2) = 5(2^{2})+(3)(2)+1-5(f(2))$ To solve for the value of $h$ , we need to solve for the value of $f(2)$ $f(2) = -4(2^{2})+(-6)(2)$ $f(2) = -28$ That means $h(2) = 5(2^{2})+(3)(2)+1+(-5)(-28)$ $h(2) = 167$